Integrand size = 21, antiderivative size = 166 \[ \int x^2 (A+B x) \left (a+b x+c x^2\right )^3 \, dx=\frac {1}{3} a^3 A x^3+\frac {1}{4} a^2 (3 A b+a B) x^4+\frac {3}{5} a \left (a b B+A \left (b^2+a c\right )\right ) x^5+\frac {1}{6} \left (3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )\right ) x^6+\frac {1}{7} \left (b^3 B+3 A b^2 c+6 a b B c+3 a A c^2\right ) x^7+\frac {3}{8} c \left (b^2 B+A b c+a B c\right ) x^8+\frac {1}{9} c^2 (3 b B+A c) x^9+\frac {1}{10} B c^3 x^{10} \] Output:
1/3*a^3*A*x^3+1/4*a^2*(3*A*b+B*a)*x^4+3/5*a*(a*b*B+A*(a*c+b^2))*x^5+1/6*(3 *a*B*(a*c+b^2)+A*(6*a*b*c+b^3))*x^6+1/7*(3*A*a*c^2+3*A*b^2*c+6*B*a*b*c+B*b ^3)*x^7+3/8*c*(A*b*c+B*a*c+B*b^2)*x^8+1/9*c^2*(A*c+3*B*b)*x^9+1/10*B*c^3*x ^10
Time = 0.05 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.00 \[ \int x^2 (A+B x) \left (a+b x+c x^2\right )^3 \, dx=\frac {1}{3} a^3 A x^3+\frac {1}{4} a^2 (3 A b+a B) x^4+\frac {3}{5} a \left (a b B+A \left (b^2+a c\right )\right ) x^5+\frac {1}{6} \left (3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )\right ) x^6+\frac {1}{7} \left (b^3 B+3 A b^2 c+6 a b B c+3 a A c^2\right ) x^7+\frac {3}{8} c \left (b^2 B+A b c+a B c\right ) x^8+\frac {1}{9} c^2 (3 b B+A c) x^9+\frac {1}{10} B c^3 x^{10} \] Input:
Integrate[x^2*(A + B*x)*(a + b*x + c*x^2)^3,x]
Output:
(a^3*A*x^3)/3 + (a^2*(3*A*b + a*B)*x^4)/4 + (3*a*(a*b*B + A*(b^2 + a*c))*x ^5)/5 + ((3*a*B*(b^2 + a*c) + A*(b^3 + 6*a*b*c))*x^6)/6 + ((b^3*B + 3*A*b^ 2*c + 6*a*b*B*c + 3*a*A*c^2)*x^7)/7 + (3*c*(b^2*B + A*b*c + a*B*c)*x^8)/8 + (c^2*(3*b*B + A*c)*x^9)/9 + (B*c^3*x^10)/10
Time = 0.45 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {1195, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^2 (A+B x) \left (a+b x+c x^2\right )^3 \, dx\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \int \left (a^3 A x^2+a^2 x^3 (a B+3 A b)+3 c x^7 \left (a B c+A b c+b^2 B\right )+3 a x^4 \left (A \left (a c+b^2\right )+a b B\right )+x^6 \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+x^5 \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+c^2 x^8 (A c+3 b B)+B c^3 x^9\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} a^3 A x^3+\frac {1}{4} a^2 x^4 (a B+3 A b)+\frac {3}{8} c x^8 \left (a B c+A b c+b^2 B\right )+\frac {3}{5} a x^5 \left (A \left (a c+b^2\right )+a b B\right )+\frac {1}{7} x^7 \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\frac {1}{6} x^6 \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+\frac {1}{9} c^2 x^9 (A c+3 b B)+\frac {1}{10} B c^3 x^{10}\) |
Input:
Int[x^2*(A + B*x)*(a + b*x + c*x^2)^3,x]
Output:
(a^3*A*x^3)/3 + (a^2*(3*A*b + a*B)*x^4)/4 + (3*a*(a*b*B + A*(b^2 + a*c))*x ^5)/5 + ((3*a*B*(b^2 + a*c) + A*(b^3 + 6*a*b*c))*x^6)/6 + ((b^3*B + 3*A*b^ 2*c + 6*a*b*B*c + 3*a*A*c^2)*x^7)/7 + (3*c*(b^2*B + A*b*c + a*B*c)*x^8)/8 + (c^2*(3*b*B + A*c)*x^9)/9 + (B*c^3*x^10)/10
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x ] && IGtQ[p, 0]
Time = 1.02 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.02
method | result | size |
norman | \(\frac {B \,c^{3} x^{10}}{10}+\left (\frac {1}{9} A \,c^{3}+\frac {1}{3} B b \,c^{2}\right ) x^{9}+\left (\frac {3}{8} A b \,c^{2}+\frac {3}{8} B a \,c^{2}+\frac {3}{8} B \,b^{2} c \right ) x^{8}+\left (\frac {3}{7} A a \,c^{2}+\frac {3}{7} A \,b^{2} c +\frac {6}{7} B a b c +\frac {1}{7} B \,b^{3}\right ) x^{7}+\left (A a b c +\frac {1}{6} A \,b^{3}+\frac {1}{2} B \,a^{2} c +\frac {1}{2} B a \,b^{2}\right ) x^{6}+\left (\frac {3}{5} a^{2} A c +\frac {3}{5} A a \,b^{2}+\frac {3}{5} B \,a^{2} b \right ) x^{5}+\left (\frac {3}{4} A \,a^{2} b +\frac {1}{4} B \,a^{3}\right ) x^{4}+\frac {a^{3} A \,x^{3}}{3}\) | \(169\) |
orering | \(\frac {x^{3} \left (252 B \,c^{3} x^{7}+280 A \,c^{3} x^{6}+840 B b \,c^{2} x^{6}+945 A b \,c^{2} x^{5}+945 B a \,c^{2} x^{5}+945 B \,b^{2} c \,x^{5}+1080 A a \,c^{2} x^{4}+1080 A \,b^{2} c \,x^{4}+2160 B a b c \,x^{4}+360 B \,b^{3} x^{4}+2520 A a b c \,x^{3}+420 A \,b^{3} x^{3}+1260 B \,a^{2} c \,x^{3}+1260 B a \,b^{2} x^{3}+1512 A \,a^{2} c \,x^{2}+1512 A a \,b^{2} x^{2}+1512 B \,a^{2} b \,x^{2}+1890 A \,a^{2} b x +630 B \,a^{3} x +840 a^{3} A \right )}{2520}\) | \(192\) |
gosper | \(\frac {1}{10} B \,c^{3} x^{10}+\frac {1}{9} x^{9} A \,c^{3}+\frac {1}{3} x^{9} B b \,c^{2}+\frac {3}{8} x^{8} A b \,c^{2}+\frac {3}{8} x^{8} B a \,c^{2}+\frac {3}{8} x^{8} B \,b^{2} c +\frac {3}{7} x^{7} A a \,c^{2}+\frac {3}{7} x^{7} A \,b^{2} c +\frac {6}{7} x^{7} B a b c +\frac {1}{7} B \,b^{3} x^{7}+x^{6} A a b c +\frac {1}{6} A \,b^{3} x^{6}+\frac {1}{2} x^{6} B \,a^{2} c +\frac {1}{2} x^{6} B a \,b^{2}+\frac {3}{5} x^{5} a^{2} A c +\frac {3}{5} x^{5} A a \,b^{2}+\frac {3}{5} x^{5} B \,a^{2} b +\frac {3}{4} x^{4} A \,a^{2} b +\frac {1}{4} a^{3} B \,x^{4}+\frac {1}{3} a^{3} A \,x^{3}\) | \(193\) |
risch | \(\frac {1}{10} B \,c^{3} x^{10}+\frac {1}{9} x^{9} A \,c^{3}+\frac {1}{3} x^{9} B b \,c^{2}+\frac {3}{8} x^{8} A b \,c^{2}+\frac {3}{8} x^{8} B a \,c^{2}+\frac {3}{8} x^{8} B \,b^{2} c +\frac {3}{7} x^{7} A a \,c^{2}+\frac {3}{7} x^{7} A \,b^{2} c +\frac {6}{7} x^{7} B a b c +\frac {1}{7} B \,b^{3} x^{7}+x^{6} A a b c +\frac {1}{6} A \,b^{3} x^{6}+\frac {1}{2} x^{6} B \,a^{2} c +\frac {1}{2} x^{6} B a \,b^{2}+\frac {3}{5} x^{5} a^{2} A c +\frac {3}{5} x^{5} A a \,b^{2}+\frac {3}{5} x^{5} B \,a^{2} b +\frac {3}{4} x^{4} A \,a^{2} b +\frac {1}{4} a^{3} B \,x^{4}+\frac {1}{3} a^{3} A \,x^{3}\) | \(193\) |
parallelrisch | \(\frac {1}{10} B \,c^{3} x^{10}+\frac {1}{9} x^{9} A \,c^{3}+\frac {1}{3} x^{9} B b \,c^{2}+\frac {3}{8} x^{8} A b \,c^{2}+\frac {3}{8} x^{8} B a \,c^{2}+\frac {3}{8} x^{8} B \,b^{2} c +\frac {3}{7} x^{7} A a \,c^{2}+\frac {3}{7} x^{7} A \,b^{2} c +\frac {6}{7} x^{7} B a b c +\frac {1}{7} B \,b^{3} x^{7}+x^{6} A a b c +\frac {1}{6} A \,b^{3} x^{6}+\frac {1}{2} x^{6} B \,a^{2} c +\frac {1}{2} x^{6} B a \,b^{2}+\frac {3}{5} x^{5} a^{2} A c +\frac {3}{5} x^{5} A a \,b^{2}+\frac {3}{5} x^{5} B \,a^{2} b +\frac {3}{4} x^{4} A \,a^{2} b +\frac {1}{4} a^{3} B \,x^{4}+\frac {1}{3} a^{3} A \,x^{3}\) | \(193\) |
default | \(\frac {B \,c^{3} x^{10}}{10}+\frac {\left (A \,c^{3}+3 B b \,c^{2}\right ) x^{9}}{9}+\frac {\left (3 A b \,c^{2}+B \left (a \,c^{2}+2 b^{2} c +c \left (2 a c +b^{2}\right )\right )\right ) x^{8}}{8}+\frac {\left (A \left (a \,c^{2}+2 b^{2} c +c \left (2 a c +b^{2}\right )\right )+B \left (4 a b c +b \left (2 a c +b^{2}\right )\right )\right ) x^{7}}{7}+\frac {\left (A \left (4 a b c +b \left (2 a c +b^{2}\right )\right )+B \left (a \left (2 a c +b^{2}\right )+2 a \,b^{2}+a^{2} c \right )\right ) x^{6}}{6}+\frac {\left (A \left (a \left (2 a c +b^{2}\right )+2 a \,b^{2}+a^{2} c \right )+3 B \,a^{2} b \right ) x^{5}}{5}+\frac {\left (3 A \,a^{2} b +B \,a^{3}\right ) x^{4}}{4}+\frac {a^{3} A \,x^{3}}{3}\) | \(226\) |
Input:
int(x^2*(B*x+A)*(c*x^2+b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
1/10*B*c^3*x^10+(1/9*A*c^3+1/3*B*b*c^2)*x^9+(3/8*A*b*c^2+3/8*B*a*c^2+3/8*B *b^2*c)*x^8+(3/7*A*a*c^2+3/7*A*b^2*c+6/7*B*a*b*c+1/7*B*b^3)*x^7+(A*a*b*c+1 /6*A*b^3+1/2*B*a^2*c+1/2*B*a*b^2)*x^6+(3/5*a^2*A*c+3/5*A*a*b^2+3/5*B*a^2*b )*x^5+(3/4*A*a^2*b+1/4*B*a^3)*x^4+1/3*a^3*A*x^3
Time = 0.07 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.00 \[ \int x^2 (A+B x) \left (a+b x+c x^2\right )^3 \, dx=\frac {1}{10} \, B c^{3} x^{10} + \frac {1}{9} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{9} + \frac {3}{8} \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{8} + \frac {1}{7} \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{7} + \frac {1}{3} \, A a^{3} x^{3} + \frac {1}{6} \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{6} + \frac {3}{5} \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{5} + \frac {1}{4} \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{4} \] Input:
integrate(x^2*(B*x+A)*(c*x^2+b*x+a)^3,x, algorithm="fricas")
Output:
1/10*B*c^3*x^10 + 1/9*(3*B*b*c^2 + A*c^3)*x^9 + 3/8*(B*b^2*c + (B*a + A*b) *c^2)*x^8 + 1/7*(B*b^3 + 3*A*a*c^2 + 3*(2*B*a*b + A*b^2)*c)*x^7 + 1/3*A*a^ 3*x^3 + 1/6*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*x^6 + 3/5*(B*a^2*b + A*a*b^2 + A*a^2*c)*x^5 + 1/4*(B*a^3 + 3*A*a^2*b)*x^4
Time = 0.03 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.21 \[ \int x^2 (A+B x) \left (a+b x+c x^2\right )^3 \, dx=\frac {A a^{3} x^{3}}{3} + \frac {B c^{3} x^{10}}{10} + x^{9} \left (\frac {A c^{3}}{9} + \frac {B b c^{2}}{3}\right ) + x^{8} \cdot \left (\frac {3 A b c^{2}}{8} + \frac {3 B a c^{2}}{8} + \frac {3 B b^{2} c}{8}\right ) + x^{7} \cdot \left (\frac {3 A a c^{2}}{7} + \frac {3 A b^{2} c}{7} + \frac {6 B a b c}{7} + \frac {B b^{3}}{7}\right ) + x^{6} \left (A a b c + \frac {A b^{3}}{6} + \frac {B a^{2} c}{2} + \frac {B a b^{2}}{2}\right ) + x^{5} \cdot \left (\frac {3 A a^{2} c}{5} + \frac {3 A a b^{2}}{5} + \frac {3 B a^{2} b}{5}\right ) + x^{4} \cdot \left (\frac {3 A a^{2} b}{4} + \frac {B a^{3}}{4}\right ) \] Input:
integrate(x**2*(B*x+A)*(c*x**2+b*x+a)**3,x)
Output:
A*a**3*x**3/3 + B*c**3*x**10/10 + x**9*(A*c**3/9 + B*b*c**2/3) + x**8*(3*A *b*c**2/8 + 3*B*a*c**2/8 + 3*B*b**2*c/8) + x**7*(3*A*a*c**2/7 + 3*A*b**2*c /7 + 6*B*a*b*c/7 + B*b**3/7) + x**6*(A*a*b*c + A*b**3/6 + B*a**2*c/2 + B*a *b**2/2) + x**5*(3*A*a**2*c/5 + 3*A*a*b**2/5 + 3*B*a**2*b/5) + x**4*(3*A*a **2*b/4 + B*a**3/4)
Time = 0.05 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.00 \[ \int x^2 (A+B x) \left (a+b x+c x^2\right )^3 \, dx=\frac {1}{10} \, B c^{3} x^{10} + \frac {1}{9} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{9} + \frac {3}{8} \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{8} + \frac {1}{7} \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{7} + \frac {1}{3} \, A a^{3} x^{3} + \frac {1}{6} \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{6} + \frac {3}{5} \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{5} + \frac {1}{4} \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{4} \] Input:
integrate(x^2*(B*x+A)*(c*x^2+b*x+a)^3,x, algorithm="maxima")
Output:
1/10*B*c^3*x^10 + 1/9*(3*B*b*c^2 + A*c^3)*x^9 + 3/8*(B*b^2*c + (B*a + A*b) *c^2)*x^8 + 1/7*(B*b^3 + 3*A*a*c^2 + 3*(2*B*a*b + A*b^2)*c)*x^7 + 1/3*A*a^ 3*x^3 + 1/6*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*x^6 + 3/5*(B*a^2*b + A*a*b^2 + A*a^2*c)*x^5 + 1/4*(B*a^3 + 3*A*a^2*b)*x^4
Time = 0.20 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.16 \[ \int x^2 (A+B x) \left (a+b x+c x^2\right )^3 \, dx=\frac {1}{10} \, B c^{3} x^{10} + \frac {1}{3} \, B b c^{2} x^{9} + \frac {1}{9} \, A c^{3} x^{9} + \frac {3}{8} \, B b^{2} c x^{8} + \frac {3}{8} \, B a c^{2} x^{8} + \frac {3}{8} \, A b c^{2} x^{8} + \frac {1}{7} \, B b^{3} x^{7} + \frac {6}{7} \, B a b c x^{7} + \frac {3}{7} \, A b^{2} c x^{7} + \frac {3}{7} \, A a c^{2} x^{7} + \frac {1}{2} \, B a b^{2} x^{6} + \frac {1}{6} \, A b^{3} x^{6} + \frac {1}{2} \, B a^{2} c x^{6} + A a b c x^{6} + \frac {3}{5} \, B a^{2} b x^{5} + \frac {3}{5} \, A a b^{2} x^{5} + \frac {3}{5} \, A a^{2} c x^{5} + \frac {1}{4} \, B a^{3} x^{4} + \frac {3}{4} \, A a^{2} b x^{4} + \frac {1}{3} \, A a^{3} x^{3} \] Input:
integrate(x^2*(B*x+A)*(c*x^2+b*x+a)^3,x, algorithm="giac")
Output:
1/10*B*c^3*x^10 + 1/3*B*b*c^2*x^9 + 1/9*A*c^3*x^9 + 3/8*B*b^2*c*x^8 + 3/8* B*a*c^2*x^8 + 3/8*A*b*c^2*x^8 + 1/7*B*b^3*x^7 + 6/7*B*a*b*c*x^7 + 3/7*A*b^ 2*c*x^7 + 3/7*A*a*c^2*x^7 + 1/2*B*a*b^2*x^6 + 1/6*A*b^3*x^6 + 1/2*B*a^2*c* x^6 + A*a*b*c*x^6 + 3/5*B*a^2*b*x^5 + 3/5*A*a*b^2*x^5 + 3/5*A*a^2*c*x^5 + 1/4*B*a^3*x^4 + 3/4*A*a^2*b*x^4 + 1/3*A*a^3*x^3
Time = 10.22 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.01 \[ \int x^2 (A+B x) \left (a+b x+c x^2\right )^3 \, dx=x^6\,\left (\frac {B\,c\,a^2}{2}+\frac {B\,a\,b^2}{2}+A\,c\,a\,b+\frac {A\,b^3}{6}\right )+x^7\,\left (\frac {B\,b^3}{7}+\frac {3\,A\,b^2\,c}{7}+\frac {6\,B\,a\,b\,c}{7}+\frac {3\,A\,a\,c^2}{7}\right )+x^4\,\left (\frac {B\,a^3}{4}+\frac {3\,A\,b\,a^2}{4}\right )+x^9\,\left (\frac {A\,c^3}{9}+\frac {B\,b\,c^2}{3}\right )+x^5\,\left (\frac {3\,B\,a^2\,b}{5}+\frac {3\,A\,c\,a^2}{5}+\frac {3\,A\,a\,b^2}{5}\right )+x^8\,\left (\frac {3\,B\,b^2\,c}{8}+\frac {3\,A\,b\,c^2}{8}+\frac {3\,B\,a\,c^2}{8}\right )+\frac {A\,a^3\,x^3}{3}+\frac {B\,c^3\,x^{10}}{10} \] Input:
int(x^2*(A + B*x)*(a + b*x + c*x^2)^3,x)
Output:
x^6*((A*b^3)/6 + (B*a*b^2)/2 + (B*a^2*c)/2 + A*a*b*c) + x^7*((B*b^3)/7 + ( 3*A*a*c^2)/7 + (3*A*b^2*c)/7 + (6*B*a*b*c)/7) + x^4*((B*a^3)/4 + (3*A*a^2* b)/4) + x^9*((A*c^3)/9 + (B*b*c^2)/3) + x^5*((3*A*a*b^2)/5 + (3*A*a^2*c)/5 + (3*B*a^2*b)/5) + x^8*((3*A*b*c^2)/8 + (3*B*a*c^2)/8 + (3*B*b^2*c)/8) + (A*a^3*x^3)/3 + (B*c^3*x^10)/10
Time = 0.22 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.81 \[ \int x^2 (A+B x) \left (a+b x+c x^2\right )^3 \, dx=\frac {x^{3} \left (252 b \,c^{3} x^{7}+280 a \,c^{3} x^{6}+840 b^{2} c^{2} x^{6}+1890 a b \,c^{2} x^{5}+945 b^{3} c \,x^{5}+1080 a^{2} c^{2} x^{4}+3240 a \,b^{2} c \,x^{4}+360 b^{4} x^{4}+3780 a^{2} b c \,x^{3}+1680 a \,b^{3} x^{3}+1512 a^{3} c \,x^{2}+3024 a^{2} b^{2} x^{2}+2520 a^{3} b x +840 a^{4}\right )}{2520} \] Input:
int(x^2*(B*x+A)*(c*x^2+b*x+a)^3,x)
Output:
(x**3*(840*a**4 + 2520*a**3*b*x + 1512*a**3*c*x**2 + 3024*a**2*b**2*x**2 + 3780*a**2*b*c*x**3 + 1080*a**2*c**2*x**4 + 1680*a*b**3*x**3 + 3240*a*b**2 *c*x**4 + 1890*a*b*c**2*x**5 + 280*a*c**3*x**6 + 360*b**4*x**4 + 945*b**3* c*x**5 + 840*b**2*c**2*x**6 + 252*b*c**3*x**7))/2520